T
E
X
source
DERIVE source
The length of a Be\'zier curve is an elliptic integral. This is a Be\'zier curve.
x
=
a
3
t
3
+
a
2
t
2
+
a
1
t
+
x
0
y
=
b
3
t
3
+
b
2
t
2
+
b
1
t
+
y
0
The control points of a Be\'zier curve are:
(
x
0
,
y
0
) (
x
1
,
y
1
) (
x
2
,
y
2
) (
x
3
,
y
3
)
This is the curve for those control points.
a
1
=3 (
x
1
−
x
0
)
a
2
=3 (
x
0
−2
x
1
+
x
2
)
a
3
=−
x
0
+3
x
1
−3
x
2
+
x
3
b
1
=3 (
y
1
−
y
0
)
b
2
=3 (
y
0
−2
y
1
+
y
2
)
b
3
=−
y
0
+3
y
1
−3
y
2
+
y
3
The length of the Be\'zier curve from 0 to
u
is:
⌠
⌡
u
0
√
(
a
1
+2
a
2
t
+3
a
3
t
2
)
2
+(
b
1
+2
b
2
t
+3
b
3
t
2
)
2
dt
The control points of an example Be\'zier curve are:
(
x
0
=6,
y
0
=8) (
x
1
=1,
y
1
=10) (
x
2
=7,
y
2
=3) (
x
3
=4,
y
3
=4)
Solve for
a
1
,
a
2
,
a
3
,
b
1
,
b
2
,
b
3
.
a
1
=−15
a
2
=33
a
3
=−20
b
1
=6
b
2
=−27
b
3
=17
The Be\'zier curve for this example is:
x
=−20
t
3
+33
t
2
−15
t
+6
y
=17
t
3
−27
t
2
+6
t
+8
For this example the length is:
3
⌠
⌡
1
0
√
689
t
4
−1492
t
3
+1076
t
2
−292
t
+29
dt
Numerical integration gives 7.237223368328592885826619210956022413967067986017504163362886516
DERIVE factored the polynomial:
689
t
4
−1492
t
3
+1076
t
2
−292
t
+29=
689
⎛
⎜
⎝
t
+
⎛
√
√
[ˉ6005]
1378
+
23377
474721
−
373
689
+
i
⎛
⎜
⎝
⎛
√
√{6005}
1378
−
23377
474721
+
7
689
⎞
⎟
⎠
⎞
⎟
⎠
⎛
⎜
⎝
t
+
⎛
√
√
[ˉ6005]
1378
+
23377
474721
−
373
689
−
i
⎛
⎜
⎝
⎛
√
√{6005}
1378
−
23377
474721
+
7
689
⎞
⎟
⎠
⎞
⎟
⎠
⎛
⎜
⎝
t
−
⎛
√
√
[ˉ6005]
1378
+
23377
474721
−
373
689
+
i
⎛
⎜
⎝
⎛
√
√{6005}
1378
−
23377
474721
−
7
689
⎞
⎟
⎠
⎞
⎟
⎠
⎛
⎜
⎝
t
−
⎛
√
√
[ˉ6005]
1378
+
23377
474721
−
373
689
+
i
⎛
⎜
⎝
7
689
−
⎛
√
√{6005}
1378
−
23377
474721
⎞
⎟
⎠
⎞
⎟
⎠
Now define values
k
,
m
,
a
,
b
:
k
=3
√
689
m
=[1,1,1,1]
b
=[−1,−1,−1,−1]
e
=[
e
1
,
e
2
,
e
3
,
e
4
]
a
=[0.216589−0.0937743
i
,0.216589+0.0937743
i
, 0.866139−0.0734550
i
,0.866139+0.0734550
i
]
Do a partial fraction expansion.
I
(
m
)=
D
219(1,
m
,4,
a
,
b
,
e
)
I
(
m
)=
I
(4
e
1
)+(1.29909+0.375097
i
)
I
(3
e
1
)+(0.383342+0.365466
i
)
I
(2
e
1
)
−(0.0228475−0.0784923
i
)
I
(
e
1
)
Reduce
I
(4
e
1
).
R
35(0,1,4,
a
,
b
,
e
)=0
A
(2
e
1
+
e
2
+
e
3
+
e
4
)+3
I
(4
e
1
)+(3.24774+0.937743
i
)
I
(3
e
1
)+(0.766684+0.730933
i
)
I
(2
e
1
)
−(0.0342712−0.117738
i
)
I
(
e
1
)=0
A
(2
e
1
+
e
2
+
e
3
+
e
4
)=−0.138815+0.00794130
i
Substitute
I
(4
e
1
).
I
(
m
)=(0.216516+0.0625162
i
)
I
(3
e
1
)+(0.127780+0.121822
i
)
I
(2
e
1
)
−(0.0114237−0.0392461
i
)
I
(
e
1
)+0.0462716−0.00264710
i
Reduce
I
(3
e
1
).
R
35(−1,1,4,
a
,
b
,
e
)=0
A
(
e
1
+
e
2
+
e
3
+
e
4
)+2
I
(3
e
1
)+(1.94864+0.562645
i
)
I
(2
e
1
)
+(0.383342+0.365466
i
)
I
(
e
1
)−(0.0114237−0.0392461
i
)
I
(0)=0
A
(
e
1
+
e
2
+
e
3
+
e
4
)=−0.0846852
Substitute
I
(3
e
1
).
I
(
m
)=−0.0655894
I
(2
e
1
)−(0.0415000+0.0123012
i
)
I
(
e
1
)
+(0.00246348−0.00389163
i
)
I
(0)+0.0554395
Reduce
I
(2
e
1
).
R
35(−2,1,4,
a
,
b
,
e
)=0
A
(
e
2
+
e
3
+
e
4
)+
I
(2
e
1
)+(0.0114237−0.0392461
i
)
I
(−
e
1
)
+(0.649549+0.187548
i
)
I
(
e
1
)=0
A
(
e
2
+
e
3
+
e
4
)=−0.949300−0.327220
i
Substitute
I
(2
e
1
).
I
(
m
)=(0.000749279−0.00257413
i
)
I
(−
e
1
)+0.00110358
I
(
e
1
)
+(0.00246348−0.00389163
i
)
I
(0)−0.00682448−0.0214622
i
Calculate the basic integrals numerically.
I
(−
e
1
)=−17.2709+33.5142
i
I
(0)=12.0414
I
(
e
1
)=−3.86310−1.12917
i
Substitute the values of the integrals.
I
(
m
)=0.0919054
k
I
(
m
)=7.23722
That is a good match.
Jim FitzSimons Mailto:cherry@neta.com
File translated from T
E
X by
T
T
H
, version 3.88.
On 28 Aug 2010, 06:59.